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Statement I

Since \(n\) and \(n + 1\) are consecutive integers, \(n(n+1)\) is a multiple of \(2\).

Thus, \(\frac{n(n+1)}{2}\) is an integer.

Statement II

Since \(n\) and \(n + 1\) are consecutive integers, \(n(n+1)\) and \(n(n+1)(n+2)\) are multiples of \(2\).

Since \(n, n + 1\) and \(n + 2\) are three consecutive integers, \(n(n+1)(n+2)\) is a multiple of \(3\).

Thus, \(n(n+1)(n+2)\) is a multiple of \(6\), and \(\frac{n(n+1)(n+2)}{6}\) is an integer.

Statement III

Since \(n\) and \(n + 1\) are two consecutive integers, \(n(n+1)\) is a multiple of \(2\).

Similarly, \((n+2)(n+3)\) is a multiple of \(2\).

Also, either n and \(n + 2\) or \(n + 1\) and \(n + 3\) are consecutive even integers. Thus, either \((n + 1)(n+3)\) is a multiple of \(8\) or \(n(n+2)\) is a multiple of \(8\) since one of them is a multiple of \(4\).

It follows that \(n(n+1)(n+2)(n+3)\) is a multiple of \(8\), and \(\frac{n(n+1)(n+2)(n+3)}{8}\) is an integer.

Therefore, the answer is E.

Answer: E

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